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Topic: custom LiFePO4 charging circuit help (Read 9811 times)

I am trying to make my own charging circuit for my robot. It will be powered by 4 solar cells which output ~17V @ 24ma. I have two LiFePO4 cells rated 3.2V @ 1100mah. I'm suspecting I'll have to make my MCU do all the monitoring, as I can't find a suitible charging circuit anywhere.

Ok, so I've been doing more research and there's something I just don't understand.All the lithium charging pages I've seen talk about the "charge current dropping to 3%". What exactly does this mean?Is it something to look for when monitoring the battery, or is it something that is applied to the battery?

Go back and look at the charging graph in the post you linked to. The solid line is the voltage and the dashed line is the current.Charging is done in stages.Stage1: The current supplied to the battery is constant. The voltage is monitored until it the cell voltage limit is reached. Go to stage 2.Stage 2: The voltage to the battery is held constant and the current is monitored. When the current is less than 3% of the starting current the battery is charged.

Stage 3: is when the battery is left on the charger for an extended time. The charger applies a voltage near the limit for until either the current drops or the voltage gets to the limit. This keeps the battery at peak charge without over charging or over heating.

I think you are getting a little mixed up with 'monitored' and 'applied' voltage.

(this is a bit crude but should illustrate my point)Let's say you have a dead 6V battery in one hand and a charger in the other. If you measure the voltage at the battery, it will be around 3-4 V (because it's dead) and the voltage of the charger will be 6V.

Once you connect the charge, both the battery and the charger are now around 6V. In order for the charger to maintain that voltage (because the battery is going to try to "pull it down" ) it produces a current that flows into the battery.

The moral of this story is when you are charging a battery, there is only one voltage and one current (between the charger and the battery). So the graph shows this, one voltage line and one current line.

Ok, so the voltage applied is always going to be the same, but in the beginning it gets "pulled down". After the voltage reaches maximum, the current required to finish the charging decreases until it get to around 3 percent, at which the battery is fully charged.

Is that correct? If it is, what's the best way to monitor the current?

How much current the charger can source, and what voltage the battery needs to be charged too.

The first stage, the charger trying to bring up the battery to 6V, but will stop at some voltage when it hits it's current limit (this is known as constant-current mode). As the battery charges, it will be not 'pulling down' as hard, until the voltage between the two comes up to the what should be the battery voltage, and then the charger will hold at that voltage.

As a result, the battery continues to charge and not 'pull' as hard, the current will decrease. This is constant voltage mode.

The biggest obstacle i see to to what you want to do is the high voltage (17V) coming from your solar cells, and the fact that the sun does not remain constant. While i have never done something like this before, i imagine this is one way to do it:

Take a look at this circuitIt will do the CC/CV (a better name for that would be max. current and max. voltage) you need.

You could modify it for a switch mode regulator - the resistor values need changing anyway.

You won't need the 4 diodes on the left side of the circuit, that is only for AC input. And using a R]resistive regulator is a waste of energy.

Soeren, I'm assuming you know of switch mode regulators the accept an input reference for voltage output. Care to link? I have only ever found Pot adjusted ones, like then ones from dimension engineering.

The LM317 is an adjustable regulator. The circuit you provided work as a CC source because the the LM317 is electronically adjusted to maintain the correct voltage based on current. If he were to adapt this circuit to a switching regulator, he would need one that had a input pin that controlled it's output voltage like the LM317 does. I guess the solution to this is two build a switching regulator circuit, as opposed to a pre built regulator.

Here is the switching regulator, constant-current, constant-voltage two-cell Li-Ion battery charging circuit.The regulation voltage (VREG) of the LM3420 can be externally trimmed by adding a single resistor from the COMP pinto the +IN pin or from the COMP pin to the GND pin, depending on the desired trim direction. Trim adjustments upto ±10% of VREG can be realized, with only a small increase in the temperature coefficient.http://www.national.com/mpf/LM/LM3420.html

If I set it up like in my last post, will I be able to simultaneously charge the batteries and run a load? Ideally I would like the batteries to be a sort of reservoir between the solar panels and the electronics.

Edit: From looking at the datasheets for the parts I posted, it seems that the schematic you provided is just a completed circuit.

The op-amp on the top of the schematic is configure as voltage follower for voltage sensing.The op-amp on the bottom of the schematic is for current sensing, all this are part of the circuit.Please read the PDF file for more detail.

The LM317 is an adjustable regulator. The circuit you provided work as a CC source because the the LM317 is electronically adjusted to maintain the correct voltage based on current. If he were to adapt this circuit to a switching regulator, he would need one that had a input pin that controlled it's output voltage like the LM317 does. I guess the solution to this is two build a switching regulator circuit, as opposed to a pre built regulator.

To answer the last thing first, I never said anything about pre-built or not.That said (or not ), it doesn't have to be adjustable, as long as there is some means to sink the output voltage - whether it's an enable pin on a pre-built chip, a transistor quenching the oscillator on a pre-built (or not), or any other way of cutting the output for any number of cycles needed.

Logged

Regards,Søren

A rather fast and fairly heavy robot with quite large wheels needs what? A lot of power?Please remember...Engineering is based on numbers - not adjectives

No, it's called a linear regulator and it dissipates the dropped power in a semiconductor (just like the 78xx and any other linear regulators).

While 'linear' regulator is a more commonly used name for it, it can still be called a resistive regulator. The university from which I graduated houses the North American Center for Advanced Power Systems, and thus most of the classes are taught from a power perspective. When modeling the steady state analysis of a circuit with a 78xx device, it can be replaced with a resistor of a variable value, because (quote from Wikipedia) "the transistor is acting like a resistor, it will waste electrical energy by converting it to heat." Which is why it is acceptable to refer to it as a 'resistive regulator' (much like my professors did) as well as a linear regulator. One way describes it's electrical characteristics while one way describes the state the semi-conductor is in.